Question

Unpolarized light of intensity $I_0$ passes through an ideal polarizer $A$. Another identical polarizer $B$ is placed behind $A$. The intensity of light beyond $B$ is found to be $\frac{I_0}{2}$. Now, another identical polarizer $C$ is placed between $A$ and $B$. The intensity beyond $B$ is now found to be $\frac{1}{8}.$ The angle between polarizer $A$ and $C$ is,

• A. ${60}^{\circ }$
• B. $0^{\circ }$
• C. ${30}^{\circ }$
• D. ${45}^{\circ }$

Answer 1

The correct option is D ${45}^{\circ }$

According to Malu's law, $I=I_0{\cos }^2\theta$

Initially, the intensity of light emerging from the polarizers $A$ and $B$ is same. This means that the angle between their transmission axes is $0^{\circ }$.

Let the angle between the transmission axis of $A$ and $C$ is ${\theta }_1$ and that between $C$ and $B$ is ${\theta }_2$. As the angle between transmission axis of $A$ and $B$ is $0^{\circ }\Rightarrow {\theta }_1={\theta }_2$.

Now the intensity distribution will be as shown below.



For polarizer $B$,
$I=\frac{I_0}{2}{\cos }^2{\theta }_1{\cos }^2{\theta }_2$
As ${\theta }_1={\theta }_2=\theta \text{(say)}$

$\therefore \frac{I_0}{8}=\frac{I_0}{2}({\cos }^2\theta {)}^2$
$\frac{1}{8}=\frac{1}{2}{\cos }^4\theta$
$\therefore \cos \theta =\frac{1}{\sqrt{2}}$
$\Rightarrow \theta ={45}^{\circ }$

Hence, $\left(D\right)$ is the correct answer.